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Home/ Questions/Q 8263985
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T04:17:46+00:00 2026-06-08T04:17:46+00:00

I am trying to make a page that displays articles in a database but

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I am trying to make a page that displays articles in a database but it is to be organised in by category.

The categories are stored in a table categories with id and category as its fields/
The articles are stored in a table articles with its different fields for various information about the article.

The code I have:

<?php

$sql = "SELECT * FROM `categories`";
$query = mysql_query($sql) or die("Could not get CATEGORIES ".mysql_error());

while($category = mysql_fetch_array($query)){

$cat = $category['category'];

$sql = "SELECT * FROM `articles` WHERE `category` = '$cat'";
$query = mysql_query($sql) or die(mysql_error());
$articles = mysql_fetch_array($query);

$size = count($articles);

echo $size;

}   

?>

Results that I expect are a list of each category, then underneath each category, the number of articles with that category.

Some help would be much appreciated.

Thanks 😀

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-08T04:17:48+00:00Added an answer on June 8, 2026 at 4:17 am

    Try this one… may be this is what you are looking for. We can also find the same result using single mysql query.

    $sql = "SELECT * FROM `categories`";
    $query = mysql_query($sql) or die("Could not get CATEGORIES ".mysql_error());
    
    $result = array();
    while($category = mysql_fetch_array($query)){
    
    $cat = $category['category'];
    
    $sql = "SELECT count(*) as numberOfArticles FROM `articles` WHERE `category` = '$cat'";
    $query2 = mysql_query($sql) or die(mysql_error());
    
    $articles = mysql_fetch_array($query2);
    $category['numberOfArticles'] = $articles['numberOfArticles'];
    $result[] = $category;
    
    }   
    
    print_r($result);
    

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